Helmut Jarausch wrote:
> Your model is A*sin(omega*t+alpha) where A and alpha are sought.
> Let T=(t_1,...,t_N)' and Y=(y_1,..,y_N)' your measurements (t_i,y_i)
> ( ' denotes transposition )
>
> First, A*sin(omega*t+alpha) =
> A*cos(alpha)*sin(omega*t) + A*sin(alpha)*cos(omega*t) =
> B*sin(omega*t) + D*cos(omega*t)
>
> by setting B=A*cos(alpha) and D=A*sin(alpha)
>
> Once, you have B and D, tan(alpha)= D/B A=sqrt(B^2+D^2)
This is all very true, but the equation tan(alpha)=D/B may fool you.
This may lead you to believe that alpha=arctan(D/B) is a solution, which
is not always the case. The point (B,D) may be in any of the four
quadrants of the plane. Assuming B!=0, the solutions to this equation
fall into the two classes
alpha = arctan(D/B) + 2*k*pi
and
alpha = arctan(D/B) + (2*k+1)*pi,
where k is an integer. The sign of B tells you which class gives you the
solution. If B is positive, the solutions are those in the first class.
If B is negative, the solutions are instead those in the second class.
Whithin the correct class, you may of course choose any alternative.
Then we have the case B=0. Then the sign of D determines alpha. If D is
positive, we have alpha=pi/2, and if D is negative, we have alpha=-pi/2.
Last if both B and D are zero, any alpha will do.
/MiO
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